Question:
A current carrying circular coil of radius 'R' produces magnetic field \(B_1\) at an axial point P at a distance 'x' from its centre and \(B_2\) at point Q placed at its centre respectively . If \(B_2 = 8B_1\), the value of x is
A current carrying circular coil of radius 'R' produces magnetic field \(B_1\) at an axial point P at a distance 'x' from its centre and \(B_2\) at point Q placed at its centre respectively . If \(B_2 = 8B_1\), the value of x is
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Compare the axial field formula with the centre value and solve for x.
Updated On: Oct 1, 2026
- \(3R\)
- \(\sqrt{3}\,R\)
- \(\frac{R}{2\sqrt{3}}\)
- \(\frac{2R}{\sqrt{3}}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understand the concept
On the axis of a circular coil, the field at distance \(x\) from the centre is \(B_1 = \dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}\). At the centre (\(x = 0\)) it is \(B_2 = \dfrac{\mu_0 I}{2R}\).
Step 2: Take the ratio
\[ \frac{B_2}{B_1} = \frac{(R^2 + x^2)^{3/2}}{R^3} = 8 \]
Step 3: Solve
\((R^2 + x^2)^{3/2} = 8R^3\), so \(R^2 + x^2 = (8)^{2/3}R^2 = 4R^2\).
Step 4: Result
\(x^2 = 3R^2\), so \(x = \sqrt{3}\,R\), option (B).
Final Answer:
The distance is sqrt 3 R. This is option (B). \[ \boxed{\text{(B) }\sqrt{3}R} \]
On the axis of a circular coil, the field at distance \(x\) from the centre is \(B_1 = \dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}\). At the centre (\(x = 0\)) it is \(B_2 = \dfrac{\mu_0 I}{2R}\).
Step 2: Take the ratio
\[ \frac{B_2}{B_1} = \frac{(R^2 + x^2)^{3/2}}{R^3} = 8 \]
Step 3: Solve
\((R^2 + x^2)^{3/2} = 8R^3\), so \(R^2 + x^2 = (8)^{2/3}R^2 = 4R^2\).
Step 4: Result
\(x^2 = 3R^2\), so \(x = \sqrt{3}\,R\), option (B).
Final Answer:
The distance is sqrt 3 R. This is option (B). \[ \boxed{\text{(B) }\sqrt{3}R} \]
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